Problem 16 - Analytical Chemistry

Question

For each of the following explain how you would prepare \(1.0 \mathrm{~L}\) of a solution that is \(0.10 \mathrm{M}\) in \(\mathrm{K}^{+}\). Repeat for concentrations of \(1.0 \times 10^{2}\) \(\operatorname{ppm} \mathrm{K}^{+}\) and \(1.0 \% \mathrm{w} / \mathrm{v} \mathrm{K}^{+}\) a. \(\mathrm{KCl}\) b. \(\mathrm{K}_{2} \mathrm{SO}_{4}\) c. \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\)

Step-by-Step Solution

Answer

Prepare solutions by dissolving calculated masses of KCl, K2SO4, and K3Fe(CN)6 in water to make 1.0 L for 0.10 M, 100 ppm, and 1% w/v K+ solutions.

Step 1: Understanding Molarity Concept

Molarity (M) is defined as moles of solute per liter of solution. To prepare a 0.10 M Kto \(1.0 \mathrm{~L}\) of solution. If using \(\mathrm{KCl}\), the molar mass is \(74.55 \text{g/mol}\).

Step 2: Calculate Amount Required for KCl

For \(0.10 \mathrm{M}\) solution of \(\mathrm{K}^{+}\), its concentration is \(0.10 \mathrm{M}\). For \(\mathrm{KCl}\), it is one-to-one with \(\mathrm{K}^{+}\); therefore, use \(0.10 moles\). Multiply by the molar mass: \[0.10 \times 74.55 = 7.455 \text{g of KCl}\].

Step 3: Prepare 1.0 L of 0.10 M KCl Solution

Dissolve \(7.455 \text{g of KCl}\) in enough water to make the final volume of \(1.0 \mathrm{~L}\) of solution.

Step 4: Calculate ppm for KCl

For \(1.0 \times 10^2\) ppm \(\mathrm{K}^{+}\), convert ppm to grams per liter: \(100 \text{ppm} = 0.100 \text{g/L}\). With \(\mathrm{KCl}\) containing \(74.55 \text{g/mol}\), from \(0.100 \text{g of } \mathrm{K}^{+} \approx 0.133 \text{g of KCl}\).

Step 5: Prepare 1.0 L of 100 ppm KCl Solution

Dissolve \(0.133 \text{g of KCl}\) in enough water to make \(1.0 \mathrm{~L}\) of solution.

Step 6: Calculate Weight/Volume Percentage for KCl

For \(1.0 \%\) \(\mathrm{w}/\mathrm{v}\) \(\mathrm{K}^{+}\), it means \(1.0 \text{g of K}^{+}\) per \(100 \text{mL}\) or \(10 \text{g/L}\). Therefore, \(10 \text{g of } \mathrm{K}^{+} \approx 13.4 \text{g of KCl}\).

Step 7: Prepare 1.0 L of 1.0% w/v KCl Solution

Dissolve \(13.4 \text{g of KCl}\) in enough water to make \(1.0 \mathrm{~L}\) of solution.

Step 8: Calculate Moles for K2SO4

For a solution of \(\mathrm{K}_2\mathrm{SO}_4\), there are two \(\mathrm{K}^{+}\) per mole. Therefore, for \(0.10 \mathrm{M}\) \(\mathrm{K}^{+}\), use \(0.050 \text{mol of } \mathrm{K}_2\mathrm{SO}_4\). Molar mass is \(174.26 \text{g/mol}\); calculate mass: \[0.050 \times 174.26 = 8.713 \text{g}\].

Step 9: Prepare 1.0 L of 0.10 M K2SO4 Solution

Dissolve \(8.713 \text{g of K}_2\mathrm{SO}_4\) in water to make a \(1.0 \mathrm{~L}\) solution.

Step 10: Calculate ppm for K2SO4

With \(1.0 \times 10^2\) ppm equaling \(0.100 \text{g of K}^{+}\), and \(2.0 \text{g/L of } \mathrm{K}_2\mathrm{SO}_4\) satisfies \(1.0 \times 10^2\) ppm \(\mathrm{K}^{+}\).

Step 11: Prepare 1.0 L of 100 ppm K2SO4 Solution

Dissolve \(2.0 \text{g of } \mathrm{K}_2\mathrm{SO}_4\) in water to make \(1.0 \mathrm{~L}\) of solution.

Step 12: Weight/Volume Percentage for K2SO4

For \(1\%\), use \(10 \text{g of K}^{+}/\text{L}\). Therefore, \[10 \times \frac{174.26}{2 \times 39.10} = 22.3 \text{g of } \mathrm{K}_2\mathrm{SO}_4\].

Step 13: Prepare 1.0 L of 1.0% w/v K2SO4 Solution

Dissolve \(22.3 \text{g of } \mathrm{K}_2\mathrm{SO}_4\) in a solution of \(1.0 \mathrm{~L}\).

Step 14: Calculate Amount Required for K3Fe(CN)6

With three \(\mathrm{K}^{+}\) ions in \(\mathrm{K}_3\mathrm{Fe(CN)}_6\), for \(0.10 \text{M }\mathrm{K}^{+}\), calculate using \(0.03334 \text{mol of } \mathrm{K}_3\mathrm{Fe(CN)}_6\):\[0.03334 \times 329.24 = 10.97 \text{g}\].

Step 15: Prepare 1.0 L of 0.10 M K3Fe(CN)6 Solution

Dissolve \(10.97 \text{g of } \mathrm{K}_3\mathrm{Fe(CN)}_6\) in enough water to make \(1.0 \mathrm{~L}\) solution.

Step 16: Calculate ppm for K3Fe(CN)6 Solution

Using \(1.0 \times 10^2\) ppm \(\mathrm{K}^{+}\), which equates to \(0.100 \text{g/L}\), calculate as\[0.100 \times \frac{329.24}{3 \times 39.10} \approx 0.280 \text{g of } \mathrm{K}_3\mathrm{Fe(CN)}_6\].

Step 17: Prepare 1.0 L of 100 ppm K3Fe(CN)6 Solution

Dissolve \(0.280 \text{g of } \mathrm{K}_3\mathrm{Fe(CN)}_6\) in water to make \(1.0 \mathrm{~L}\) of solution.

Step 18: Weight/Volume Percentage for K3Fe(CN)6

For \(1\%\) \(\mathrm{w}/\mathrm{v}\), convert to \[10.0 \times \frac{329.24}{3 \times 39.10} = 28.2 \text{g of } \mathrm{K}_3\mathrm{Fe(CN)}_6\].

Step 19: Prepare 1.0 L of 1.0% w/v K3Fe(CN)6 Solution

Dissolve \(28.2 \text{g of } \mathrm{K}_3\mathrm{Fe(CN)}_6\) in water to make a \(1.0 \mathrm{~L}\) solution.

Key Concepts

Molarity

Molarity is a fundamental concept in solution preparation. It helps determine how many moles of a solute are present in a liter of solution. To calculate molarity, use the formula: \[ M = \frac{n}{V} \] where \( M \) is molarity, \( n \) is the number of moles, and \( V \) is the volume in liters.
Understanding molarity is crucial for preparing solutions accurately, as it allows students to determine the exact amount of substance needed when given a target solution concentration. For instance, when preparing a 0.10 M solution of potassium ions (K\(^+\)), you need the moles based on \[ M = 0.10 \, \text{mol/L}. \]
Given the molar mass, you can compute the grams needed to achieve this concentration, ensuring a precise preparation for various compounds like KCl, K\(_2\)SO\(_4\), and K\(_3\)Fe(CN)\(_6\).

ppm (parts per million)

In chemistry, ppm, or parts per million, is a way to express very dilute concentrations of substances. It is a measurement of concentration that tells you how many parts of a solute are present per million parts of the solution. This is particularly helpful for trace amount preparations where a high degree of accuracy is necessary.
To calculate ppm: \[ \text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6 \]In the exercise, preparing a solution with 100 ppm of K\(^+\) means converting ppm into g/L, which translates to: \[ 100 \, \text{ppm} = 0.100 \, \text{g/L}. \] This conversion helps you prepare precise solutions, crucial for experiments where potassium ions play a significant role. Understanding ppm ensures students can handle both low and precisely accurate concentrations in laboratories.

Weight/Volume Percentage

Weight/volume percentage (w/v %) is a common way to express concentration in solutions, particularly in biology and chemistry labs. It is defined as the mass of solute divided by the volume of the solution, multiplied by 100. This indicator shows how much solute is present in a specific volume of solution, making it useful for creating solutions where a certain percentage is required.
For example: \[ \text{w/v \,%} = \frac{\text{weight of solute (g)}}{\text{volume of solution (mL)}} \times 100 \]In the problem scenario, preparing a solution with a 1% w/v K\(^+\) involves ensuring that every liter contains 10 g of potassium ions. Calculating the corresponding weight for the compounds allows students to prepare the exact solutions needed for laboratory work. The understanding of w/v % makes it simpler for students to visualize and measure out the necessary components for solution preparation.

Potassium Ions

Potassium ions (K\(^+\)) are frequently encountered in various chemical reactions and solutions. They are crucial for biological activities and numerous chemical processes. In solution preparation exercises, understanding the role and calculation of potassium ions helps students determine the requisite concentration in solutions.
For any compound providing potassium ions, it is important to consider the number of potassium ions contributed by one mole of the compound. For instance:

• KCl provides one K\(^+\)
• K\(_2\)SO\(_4\) provides two K\(^+\)
• K\(_3\)Fe(CN)\(_6\) provides three K\(^+\)

Recognizing the stoichiometry or the ratio in which potassium ions are provided enables the calculation of the mass of the compound needed for a specific molarity or ppm solution. This understanding is vital for students practicing solution preparation in compounds involving potassium ions.

Chemical Calculations

Chemical calculations are pivotal in determining how much of a particular substance is needed to prepare a desired solution. These calculations often involve the conversion of units and careful application of stoichiometric principles. The calculation process ensures that the right amounts of chemicals are used, avoiding wastage or inaccurately prepared solutions.
Here are some essential steps for chemical calculations in solution preparation:

• Determine the molarity, ppm, or percentage needed for the solution.
• Calculate the moles required based on the concentration and volume.
• Use the molar mass to convert moles into the mass of the compound.
• Adjust calculations for compounds that yield multiple ions (like potassium ions).

Careful and clear chemical calculations give students confidence in their preparation of solutions and help reinforce critical thinking skills needed in chemical lab environments. These skills are not only important for student exercises but are also vital in professional chemical industries.